How do you integrate int x^2e^(x^3/2)dx from [-2,3]?
1 Answer
Jan 14, 2017
Explanation:
I=int_(-2)^3x^2e^(x^3/2)dx
Use the substitution:
u=x^3/2" "=>" "du=3/2x^2dx
We already have
x=3" "=>" "u=3^3/2=27/2
x=-2" "=>" "u=(-2)^3/2=-4
Thus:
I=2/3inte^(x^3/2)(3/2x^2dx)=2/3int_(-4)^(27/2)e^udu
The integral of
=2/3[e^u]_(-4)^(27/2)=2/3(e^(27/2)-e^(-4))=2/3((e^(35/2)-1)/e^4)=(2e^(35/2)-2)/(3e^4)